a, (-0.25) * 7.9 * 40 = [ (-0.25) * 40] * 7.9 = -10 * 7.9 = -79

b, ( 3/2 )^3 * 2^3 = ( 3/2*2) ^3 = 3^3 = 27

c, (1.75 / \(\frac{7}{2}\)) * \(\frac{4}{5}\)= ( 7/4 * 7/2 ) *4/5 = \(\frac{49}{10}\)

d, \(\frac{11}{2}\cdot4\frac{5}{3}-2\frac{5}{3}\cdot\frac{11}{2}\)= \(\frac{11}{2}\cdot\left(\frac{17}{3}-\frac{11}{3}\right)\)= \(\frac{11}{2}\cdot\frac{6}{3}\)= \(\frac{11}{2}\cdot2\)=11

ẽ2d3z3

a: \(=\left(1.25\right)^{16}\cdot8^{16}\cdot8=8\cdot10^{16}\)

b: \(=\left(\dfrac{5}{2}\right)^{13}\cdot4^{13}\cdot4^2=10^{13}\cdot4^2\)

c: \(=\left(0.25\right)^4\cdot8^4\cdot8^2=2^4\cdot8^2=64\cdot16=1024\)

d: \(=\left(\dfrac{1}{2}\right)^{15}\cdot2^{18}=2^3=8\)

e: \(=\left(\dfrac{1}{3}\cdot6\right)^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=2^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=\dfrac{1}{2}\)

\(a;0,25-\frac{1}{2}\left|1,5-x\right|=2,5\)

\(\Leftrightarrow\frac{1}{2}\left|1,5-x\right|=0,25-2,5\)

\(\Leftrightarrow\frac{1}{2}\left|1,5-x\right|=-2,25\)

\(\Leftrightarrow\left|1,5-x\right|=-2,25\cdot2=-4,5\)

Mà \(\left|1,5-x\right|\ge0\)Nên suy ra |1,5-x|=-4,5 là vô lý

\(b;\left|x+\frac{1}{6}\right|\cdot0,75+\frac{1}{4}=2\frac{1}{3}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|\cdot\frac{3}{4}=\frac{7}{3}-\frac{1}{4}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|\cdot\frac{3}{4}=\frac{25}{12}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|=\frac{25}{12}\cdot\frac{4}{3}\)

\(\Leftrightarrow\left|x+\frac{1}{6}\right|=\frac{25}{9}\Leftrightarrow x+\frac{1}{6}=\pm\frac{25}{9}\)

TH1:\(x+\frac{1}{6}=\frac{25}{9}\)

\(\Leftrightarrow x=\frac{25}{9}-\frac{1}{6}=\frac{47}{18}\)

TH2:\(x+\frac{1}{6}=-\frac{25}{9}\)

\(\Leftrightarrow x=-\frac{25}{9}-\frac{1}{6}=\frac{-53}{18}\)

Vậy \(x=\frac{47}{18};-\frac{53}{18}\)